Question

As shown in the figure, water squirts horizontally out of two small holes in the side of cylinder and the two streams strike the ground at the same point. If the level of water stands at height $H=4.25\text{m}$ and hole $P$ is at depth of $0.25\text{m}$ below the surface of water, then find the depth of hole $Q$ from the surface of water.

• A. $3\text{m}$
• B. $2\text{m}$
• C. $0.5\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is D $4\text{m}$


We know,
Range of fluid $R=2\sqrt{h(H-h)}$
So, for point $P$,
$R=2\sqrt{h(H-h)}$
For point $Q$,
$R^{\prime }=2\sqrt{h^{\prime }(H-h^{\prime })}$
Given,
$R=R^{\prime }$
$\therefore 2\sqrt{h(H-h)}=2\sqrt{h^{\prime }(H-h^{\prime })}$
$\Rightarrow h^{\prime 2}-h^{\prime }H+h(H-h)=0$
$\Rightarrow h^{\prime 2}-4.25h^{\prime }+0.25(4.25-0.25)=0$
$\Rightarrow h^{\prime 2}-4.25h^{\prime }+1=0$
$\Rightarrow (h^{\prime }-4)(h^{\prime }-\frac{1}{4})=0$
$\Rightarrow h^{\prime }=4$ or $0.25$
$\therefore$ The depth of $Q$ from the surface of water is $h^{\prime }=4\text{m}$